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calculate the L of 5.00 M HCI required to neutralize 500.0 mL of 1.00 M Ca

  • chemistry -

    Ca(OH)2 + 2HCl = CaCl2 + 2H2O
    So, each mole of Ca(OH)2 neutralizes 2 moles of HCl

    500mL of 1.00M Ca(OH)2 = 0.50 moles
    so, you need 1.00 moles of HCl
    That would be 200mL of 5.0M acid

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